The proof of this theorem is as follows. The Monotone convergence axiom states that any bounded monotonic
sequence of real numbers converges, so it only needs to be shown that any sequence of real numbers has a
monotonic subsequence.

Let (an) be a sequence of real numbers. Say that k is a maximal index if
ak>=al for all l>=k. Then there are two possibilities:

There are infinitely many maximal indices k1<k2<k3<...
Then ak_1>=ak_2>=ak_3>... so (an) has a decreasing
subsequence.

There are finitely many maximal indices. Let b be an upper bound for the set of maximal indices. Then if
n>b, n isn't a maximal index. It can be shown by induction that there exists an increasing subsequence
(ak_j). Let k1=b+1. k1 isn't a maximal index, so there exists
k2>k1 with ak_2>ak_1. Now, suppose we have found
kj. kj isn't a maximal index, so there exists kj+1>kj with
ak_j+1>ak_j. So one can find an increasing subsequence.